AP Chemistry Intermolecular Forces and Properties — Worked Answer Explanations

Unit 3 · 12 questions explained

Below is a complete answer key for our AP Chemistry Intermolecular Forces and Properties practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Intermolecular Forces and Properties practice test and come back here to review, or head back to the Intermolecular Forces and Properties unit overview.

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Question 1 · Easy
Which type of intermolecular force is present in ALL molecular substances?
- A
  Hydrogen bonding
  Why not A: Hydrogen bonding requires H bonded to N, O, or F — not present in all molecules.
- B
  Dipole–dipole forces
  Why not B: Dipole–dipole requires a net dipole moment, which nonpolar molecules lack.
- C
  London dispersion forcesCorrect
- D
  Ion–dipole forces
  Why not D: Ion–dipole forces involve ions, not just neutral molecules.
Explanation
London dispersion forces (induced dipole–induced dipole) arise from temporary fluctuations in electron distribution. Every molecule, polar or nonpolar, has electrons, so all experience LDFs. They are typically the weakest IMF but are universal.

Key takeaway
London dispersion forces are present in every molecular substance because all molecules have electrons.
Question 2 · Easy
Explain why $HF$ has a much higher boiling point ( $19.5° C$ ) than $HCl$ ( $- 85° C$ ), even though HCl has the larger molar mass.
- A
  HF has stronger London dispersion forces because fluorine is highly electronegative.
  Why not A: LDF strength depends on polarizability and molar mass, not electronegativity per se.
- B
  HF molecules engage in hydrogen bonding, a much stronger intermolecular attraction than the dipole–dipole forces in HCl.Correct
- C
  The H–F bond is covalent while H–Cl is ionic.
  Why not C: Both H–F and H–Cl bonds are polar covalent.
- D
  HF has a higher molar mass than HCl.
  Why not D: HCl (36.5 g/mol) has a higher molar mass than HF (20 g/mol).
Explanation
Fluorine is the most electronegative element; the H bonded to F is sufficiently positive to form true hydrogen bonds with the lone pairs of neighboring F atoms. These H-bonds ( $\sim 20 kJ/mol$ ) are far stronger than the dipole–dipole interactions in HCl, more than compensating for HCl's larger size and greater LDF.

Key takeaway
Hydrogen bonding (H bonded to N, O, or F) can dominate boiling-point trends over molecular mass.
Question 3 · Easy
A gas occupies $3.00 L$ at $2.00 atm$ and $300 K$ . What volume will it occupy at $1.00 atm$ and $600 K$ , assuming ideal behavior?
- A
  $3.00 L$
  Why not A: Doubling T doubles V, but halving P also doubles V — combine both effects.
- B
  $6.00 L$
  Why not B: Only one of the two changes (either P or T) was applied.
- C
  $12.0 L$ Correct
- D
  $1.50 L$
  Why not D: Incorrectly inverted one of the ratios.
Explanation
Combined gas law: $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$

$V_{2} = V_{1} \times \frac{P _{1}}{P _{2}} \times \frac{T _{2}}{T _{1}} = 3.00 \times \frac{2.00}{1.00} \times \frac{600}{300} = 3.00 \times 2 \times 2 = 12.0 L$

Key takeaway
Combined gas law: $PV/T = \text{const}$. Halving pressure doubles volume; doubling temperature doubles volume.
Question 4 · Easy
A solution is prepared by dissolving $5.85 g$ of NaCl (molar mass $58.5 g/mol$ ) in enough water to make $250 mL$ of solution. What is the molarity of NaCl?
- A
  $0.100 M$
  Why not A: Off by a factor of 4; 250 mL = 0.250 L, not 1 L.
- B
  $0.400 M$ Correct
- C
  $2.50 M$
  Why not C: Divided moles by 0.250 g instead of 0.250 L, or made a factor-of-10 error.
- D
  $0.0400 M$
  Why not D: Used 250 L instead of 250 mL.
Explanation
Moles NaCl $= \frac{5.85}{58.5} = 0.100 mol$

Molarity $= \frac{0.100 mol}{0.250 L} = 0.400 M$

Key takeaway
Molarity = moles of solute / liters of solution. Always convert mL to L.
Question 5 · Medium
At $0° C$ and $1.00 atm$ , $1 mol$ of a real gas occupies a slightly smaller volume than predicted by the ideal gas law. What does this suggest about the dominant deviation at this condition?
- A
  Molecular volume is dominant, pushing measured volume above the ideal.
  Why not A: Molecular volume increases the measured volume, not decreases it.
- B
  Intermolecular attractions are dominant, pulling molecules together and reducing volume below the ideal.Correct
- C
  The gas molecules are moving faster than predicted by the kinetic molecular theory.
  Why not C: Molecular speed affects pressure, not volume deviation directly.
- D
  The gas is undergoing a phase change at this condition.
  Why not D: A volume smaller than ideal is consistent with the van der Waals $a$ correction, not necessarily a phase change.
Explanation
For real gases, two factors cause deviations from ideal behavior. Intermolecular attractive forces draw molecules together, reducing the pressure (and effective volume) below the ideal prediction. Finite molecular volume does the opposite. At relatively low temperatures and pressures, attractions dominate, so measured volume is slightly less than ideal.

Key takeaway
Real-gas volume < ideal: intermolecular attractions dominate. Real-gas volume > ideal: finite molecular volume dominates.
Question 6 · Medium
Which of the following correctly explains why water has an unusually high surface tension?
- A
  Water molecules at the surface are attracted only by gravity, not by IMFs.
  Why not A: IMFs still act on surface molecules — they just have fewer neighbors above them.
- B
  Surface water molecules experience a net inward force due to extensive hydrogen bonding with interior molecules, resisting surface expansion.Correct
- C
  Water has a low molar mass, so surface molecules move slowly and resist disruption.
  Why not C: Low molar mass would suggest lower surface tension, not higher.
- D
  London dispersion forces in water are exceptionally strong because of water's high electronegativity.
  Why not D: Surface tension in water is due to H-bonding, not LDF; electronegativity does not directly determine LDF strength.
Explanation
Interior water molecules are surrounded on all sides by hydrogen-bonding neighbors. Surface molecules have neighbors only below and to the sides, creating a net inward pull. This cohesion force is large because H-bonds are strong, giving water one of the highest surface tensions of any common liquid.

Key takeaway
High surface tension in water arises from extensive hydrogen bonding creating a strong net inward force on surface molecules.
Question 7 · Medium
A sample of ideal gas has a volume of $10.0 L$ at $27° C$ . At constant pressure, to what temperature (in $° C$ ) must the gas be cooled to reduce its volume to $7.50 L$ ?
- A
  $- 48° C$ Correct
- B
  $0° C$
  Why not B: 273 K gives a 9.1 L volume at these conditions, not 7.50 L.
- C
  $20.25° C$
  Why not C: Used a ratio of Celsius temperatures rather than Kelvin temperatures.
- D
  $- 75° C$
  Why not D: Applied the ratio inversely.
Explanation
Charles's Law at constant $P$ : $\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$

$T_{1} = 27 + 273 = 300 K$

$T_{2} = T_{1} \times \frac{V _{2}}{V _{1}} = 300 \times \frac{7.50}{10.0} = 225 K = - 48° C$

Key takeaway
Charles's Law: $V/T$ = const at fixed $P$. Always convert to Kelvin before calculating.
Question 8 · Medium
Rank these liquids in order of increasing viscosity: ethanol ( $C_{2} H_{5} OH$ ), glycerol ( $C_{3} H_{5} (OH)_{3}$ ), and hexane ( $C_{6} H_{14}$ ).
- A
  Hexane $<$ Ethanol $<$ GlycerolCorrect
- B
  Ethanol $<$ Hexane $<$ Glycerol
  Why not B: Ethanol's H-bonding makes it more viscous than hexane (only LDF).
- C
  Glycerol $<$ Ethanol $<$ Hexane
  Why not C: Glycerol has three OH groups and extensive H-bonding — it is the most viscous.
- D
  Hexane $<$ Glycerol $<$ Ethanol
  Why not D: Glycerol's three OH groups create much stronger intermolecular interactions than ethanol's one.
Explanation
Viscosity reflects resistance to flow — driven by IMF strength and molecular entanglement. Hexane has only LDF (weakest IMFs, lowest viscosity). Ethanol has 1 OH group enabling H-bonding (intermediate viscosity). Glycerol has 3 OH groups with extensive H-bonding (highest viscosity).

Key takeaway
Greater number and strength of intermolecular forces increases viscosity.
Question 9 · Medium
At constant temperature, the pressure of a fixed amount of gas is increased from $1.00 atm$ to $4.00 atm$ . By what factor does the volume change?
- A
  Increases by a factor of 4
  Why not A: Increasing pressure at constant T decreases volume (Boyle's Law).
- B
  Decreases by a factor of 4Correct
- C
  Decreases by a factor of 2
  Why not C: Pressure quadrupled, so volume is reduced by a factor of 4, not 2.
- D
  Remains unchanged
  Why not D: Pressure and volume are inversely proportional at constant T and n (Boyle's Law).
Explanation
Boyle's Law: $P_{1} V_{1} = P_{2} V_{2}$ at constant $T$ and $n$ . If pressure quadruples ( $P_{2} = 4 P_{1}$ ), then $V_{2} = V_{1} /4$ — volume decreases by a factor of 4.

Key takeaway
Boyle's Law: $PV$ = constant at fixed $T$ and $n$. Doubling pressure halves volume.
Question 10 · Hard
According to the kinetic molecular theory of gases, what happens to the average kinetic energy of gas molecules when temperature is doubled (from 300 K to 600 K)?
- A
  It doubles.Correct
- B
  It quadruples.
  Why not B: Kinetic energy is directly proportional to T, not T $^{2}$ .
- C
  It increases by $2$ .
  Why not C: rms speed increases by $2$ , but kinetic energy depends on speed $^{2}$ , so KE doubles.
- D
  It remains the same; only pressure changes.
  Why not D: Temperature is a direct measure of average kinetic energy.
Explanation
From KMT: $\overline{K E} = \frac{3}{2} k_{B} T$ (for a monatomic ideal gas). Kinetic energy is directly proportional to absolute temperature $T$ . If $T$ doubles, $\overline{K E}$ doubles. Note: rms speed $\propto T$ , so speed increases by $2$ , but $K E \propto v^{2} \propto T$ doubles.

Key takeaway
Average kinetic energy of gas molecules is directly proportional to absolute temperature: $\overline{KE} = \tfrac{3}{2}k_BT$.
Question 11 · Hard
A 5.00-L container holds $0.500 mol$ of an ideal gas at $27° C$ . What is the pressure in atm? ( $R = 0.0821 L \cdot atm/ (mol \cdot K)$ )
- A
  $0.274 atm$
  Why not A: Used 27 K instead of 300 K for temperature.
- B
  $2.46 atm$ Correct
- C
  $4.93 atm$
  Why not C: Used 1 L instead of 5 L, or a factor-of-2 error.
- D
  $0.492 atm$
  Why not D: Forgot to multiply by the 0.500 mol factor or divided by T wrong.
Explanation
$P V = n R T \Rightarrow P = \frac{n R T}{V}$

$T = 27 + 273 = 300 K$

$P = \frac{( 0.500 ) ( 0.0821 ) ( 300 )}{5.00} = \frac{12.315}{5.00} \approx 2.46 atm$

Key takeaway
Ideal gas law: $PV = nRT$. Always use Kelvin for temperature.
Question 12 · Hard
A mixture of gas A (molar mass $4 g/mol$ ) and gas B (molar mass $64 g/mol$ ) effuses through a small orifice. According to Graham's Law, how much faster does gas A effuse compared to gas B?
- A
  4 times fasterCorrect
- B
  16 times faster
  Why not B: Used the ratio of masses directly instead of the square root of the ratio.
- C
  $4 = 2$ times faster
  Why not C: Took the square root of 4 rather than the square root of 64/4 = 16.
- D
  8 times faster
  Why not D: Computed $64/4 = 16$ and took half rather than the square root.
Explanation
Graham's Law: $\frac{r _{A}}{r _{B}} = \frac{M _{B}}{M _{A}} = \frac{64}{4} = 16 = 4$ .

Gas A (helium, $M = 4$ ) effuses 4 times faster than gas B (sulfur dioxide or similar, $M = 64$ ).

Key takeaway
Graham's Law: rate of effusion $\propto 1/\sqrt{M}$. Lighter gas effuses faster by the square-root ratio of molar masses.